L is a decidable language because the Turing machine accepts it.
L is a recognizable language if TM M recognizes it.
<h3>How do you know if a language is decidable?</h3>
A language is said to be decidable only when there seems to exists a Turing machine that is said to accepts it,
Here, it tends to halts on all inputs, and then it answers "Yes" on words that is seen in the language and says "No" on words that are not found in the language. The same scenario applies to recognizable language.
So, L is a decidable language because the Turing machine accepts it.
L is a recognizable language if TM M recognizes it.
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<span>C. Documents that convey buyers, sellers, and purchases made</span>
Answer:
Following are the response to the given question:
Explanation:
The glamorous objective is to examine the items (as being the most valuable and "cheapest" items are chosen) while no item is selectable - in other words, the loading can be reached.
Assume that such a strategy also isn't optimum, this is that there is the set of items not including one of the selfish strategy items (say, i-th item), but instead a heavy, less valuable item j, with j > i and is optimal.
As 
, the i-th item may be substituted by the j-th item, as well as the overall load is still sustainable. Moreover, because 
and this strategy is better, our total profit has dropped. Contradiction.
Rand.int(your_num , your_num
Answer:
int x;
indata.open("lottowins");
indata >> x;
cout << x << endl;
indata >> x;
cout << x << endl;
indata >> x;
cout << x << endl;
indata.close();
